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Slides ensae-2016-11

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slides ENSAE Actuariat Assurance Dommage #11

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Slides ensae-2016-11

  1. 1. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Actuariat de l’Assurance Non-Vie # 11 A. Charpentier (Université de Rennes 1) ENSAE ParisTech, Octobre / Décembre 2016. http://freakonometrics.hypotheses.org @freakonometrics 1
  2. 2. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Regression Models in Claims Reserving A natural idea is to assume that incremental payments Yi,j can be explained by two factors: one related to occurrence year i, and one development factor, related to j. Formally, we assume that Yi,j ∼ L(θi,j), where θi,j = αi · βj i.e. Yi,j is a random variable, with distribution L, where parameter(s) can be related to the two factors. @freakonometrics 2
  3. 3. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Poisson regression in claims reserving Renshaw & Verrall (1998) proposed to use a Poisson regression for incremental payments to estimate claim reserve, i.e. Yi,j ∼ P (exp [γ + αi + βj]) . 1 devF=as.factor(development); anF=as.factor(year) 2 REG=glm(vec.C~devF+anF , family = "Poisson") Here, 1 > summary(REG) 2 Call: 3 glm(formula = vec.C ~ anF + devF , family = poisson(link = "log"), 4 data = triangle) 5 6 Deviance Residuals: 7 Min 1Q Median 3Q Max 8 -2.343e+00 -4.996e -01 9.978e -07 2.770e-01 3.936e+00 @freakonometrics 3
  4. 4. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 9 10 Coefficients : 11 Estimate Std. Error z value Pr(>|z|) 12 (Intercept) 8.05697 0.01551 519.426 < 2e-16 *** 13 anF1989 0.06440 0.02090 3.081 0.00206 ** 14 anF1990 0.20242 0.02025 9.995 < 2e-16 *** 15 anF1991 0.31175 0.01980 15.744 < 2e-16 *** 16 anF1992 0.44407 0.01933 22.971 < 2e -16 *** 17 anF1993 0.50271 0.02079 24.179 < 2e -16 *** 18 devF1 -0.96513 0.01359 -70.994 < 2e -16 *** 19 devF2 -4.14853 0.06613 -62.729 < 2e-16 *** 20 devF3 -5.10499 0.12632 -40.413 < 2e -16 *** 21 devF4 -5.94962 0.24279 -24.505 < 2e-16 *** 22 devF5 -5.01244 0.21877 -22.912 < 2e -16 *** 23 --- 24 25 ( Dispersion parameter for poisson family taken to be 1) 26 @freakonometrics 4
  5. 5. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 27 Null deviance: 46695.269 on 20 degrees of freedom 28 Residual deviance: 30.214 on 10 degrees of freedom 29 AIC: 209.52 30 31 Number of Fisher Scoring iterations: 4 Again, it is possible to summarize this information in triangles.... Predictions can be used to complete the triangle. 1 ANew=rep(1 :Ntr),times=Ntr) ; DNew=rep (0 :(Ntr -1),each=Ntr) 2 P=predict(REG , newdata=data.frame(A=as.factor(ANew),D=as.factor( DNew))) 3 payinc.pred= exp(matrix(as.numeric(P),nrow=n,ncol=n)) 4 noise = payinc -payinc.pred 1 year development paycum payinc payinc.pred noise 2 1 1988 0 3209 3209 3155.699242 5.330076e+01 3 2 1989 0 3367 3367 3365.604828 1.395172e+00 4 3 1990 0 3871 3871 3863.737217 7.262783e+00 @freakonometrics 5
  6. 6. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 5 4 1991 0 4239 4239 4310.096418 -7.109642e+01 6 5 1992 0 4929 4929 4919.862296 9.137704e+00 7 6 1993 0 5217 5217 5217.000000 1.818989e-12 8 7 1988 1 4372 1163 1202.109851 -3.910985e+01 9 8 1989 1 4659 1292 1282.069808 9.930192e+00 10 9 1990 1 5345 1474 1471.824853 2.175147e+00 11 10 1991 1 5917 1678 1641.857784 3.614222e+01 12 11 1992 1 6794 1865 1874.137704 -9.137704e+00 13 12 1988 2 4411 39 49.820712 -1.082071e+01 14 13 1989 2 4696 37 53.134604 -1.613460e+01 15 14 1990 2 5398 53 60.998886 -7.998886e+00 16 15 1991 2 6020 103 68.045798 3.495420e+01 17 16 1988 3 4428 17 19.143790 -2.143790e+00 18 17 1989 3 4720 24 20.417165 3.582835e+00 19 18 1990 3 5420 22 23.439044 -1.439044e+00 20 19 1988 4 4435 7 8.226405 -1.226405e+00 21 20 1989 4 4730 10 8.773595 1.226405e+00 22 21 1988 5 4456 21 21.000000 -2.842171e-14 @freakonometrics 6
  7. 7. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 The pearson residuals are εP i,j = Xi,j − µi,j µi,j , The deviance residuals are εD i,j = Xi,j − µi,j di,j , Pearson’s error can be obtained from function resid=residuals(REG,"pearson"), and summarized in a triangle 1 > PEARSON 2 [,1] [,2] [,3] [,4] [,5] [,6] 3 [1,] 9.4882e-01 -1.128012 -1.5330 -0.48996 -0.42759 -6.2021e-15 4 [2,] 2.4048e-02 0.277333 -2.2134 0.79291 0.41404 NA 5 [3,] 1.1684e-01 0.056697 -1.0241 -0.29723 NA NA 6 [4,] -1.0829e+00 0.891963 4.2373 NA NA NA 7 [5,] 1.3027e-01 -0.211074 NA NA NA NA 8 [6,] 2.5183e-14 NA NA NA NA NA @freakonometrics 7
  8. 8. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Errors in GLMs @freakonometrics 8
  9. 9. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 log-Poisson regression and Chain-Ladder The log-Poisson is interesting since it (usually) provides the same amount of reserves as Chain Ladder. 1 > library( ChainLadder ) 2 > an <- 10; ligne = rep (1:an , each=an); colonne = rep (1:an , an) 3 > passe = (ligne + colonne - 1) <=an; n = sum(passe) 4 > PAID=GenIns; INC=PAID 5 > INC [,2:an]= PAID [,2:an]-PAID [ ,1:(an -1)] 6 > Y = as.vector(INC) 7 > lig = as.factor(ligne) 8 > col = as.factor(colonne) 9 > base = data.frame(Y,col ,lig) 10 > reg=glm(Y~col+lig ,data=base ,family="poisson") 11 > sum(exp(predict(reg ,newdata=base))[passe!=TRUE ]) 12 [1] 18680856 @freakonometrics 9
  10. 10. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 log-Poisson regression and Chain-Ladder 1 > MackChainLadder (GenIns) 2 MackChainLadder (Triangle = GenIns) 3 Latest Dev.To.Date Ultimate IBNR Mack.S.E CV(IBNR) 4 1 3 ,901 ,463 1.0000 3 ,901 ,463 0 0 NaN 5 2 5 ,339 ,085 0.9826 5 ,433 ,719 94 ,634 71 ,835 0.759 6 3 4 ,909 ,315 0.9127 5 ,378 ,826 469 ,511 119 ,474 0.254 7 4 4 ,588 ,268 0.8661 5 ,297 ,906 709 ,638 131 ,573 0.185 8 5 3 ,873 ,311 0.7973 4 ,858 ,200 984 ,889 260 ,530 0.265 9 6 3 ,691 ,712 0.7223 5 ,111 ,171 1 ,419 ,459 410 ,407 0.289 10 7 3 ,483 ,130 0.6153 5 ,660 ,771 2 ,177 ,641 557 ,796 0.256 11 8 2 ,864 ,498 0.4222 6 ,784 ,799 3 ,920 ,301 874 ,882 0.223 12 9 1 ,363 ,294 0.2416 5 ,642 ,266 4 ,278 ,972 970 ,960 0.227 13 10 344 ,014 0.0692 4 ,969 ,825 4 ,625 ,811 1 ,362 ,981 0.295 14 Totals 15 Latest: 34 ,358 ,090.00 16 Ultimate: 53 ,038 ,945.61 17 IBNR: 18 ,680 ,855.61 @freakonometrics 10
  11. 11. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 18 Mack S.E.: 2 ,441 ,364.13 19 CV(IBNR): 0.13 @freakonometrics 11
  12. 12. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 An explicit expression to quantify uncertainty Recall that we while to estimate E([R − R]2 ) = E(R) − E(R) 2 + Var(R − R) ≈ Var(R) + Var(R) Classically, consider a log-Poisson model, were incremental payments satisfy Yi,j ∼ P(µi,j) where µi,j = exp[ηi,j] = exp[γ + αi + βj] Using the delta method, we get that asymptotically Var(Yi,j) = Var(µi,j) ≈ ∂µi,j ∂ηi,j 2 Var(ηi,j) where, since we consider a log link, ∂µi,j ∂ηi,j = µi,j @freakonometrics 12
  13. 13. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 i.e., with an ODP distribution (i.e. Var(Yi,j) = ϕE(Yi,j), E [Yi,j − Yi,j]2 ≈ ϕ · µi,j + µ2 i,j · Var(ηi,j) and Cov(Yi,j, Yk,l) ≈ µi,j · µk,l · Cov (ηi,j, ηk,l) Thus, since the overall amount of reserves satisfies E [R − R]2 ≈ i+j−1>n ϕ · µi,j + µ Var(η)µ. 1 > an <- 6; ligne = rep (1:an , each=an); colonne = rep (1:an , an) 2 > passe = (ligne + colonne - 1) <=an; np = sum(passe) 3 > futur = (ligne + colonne - 1)> an; nf = sum(passe) 4 > INC=PAID 5 > INC [ ,2:6]= PAID [,2:6]- PAID [ ,1:5] 6 > Y = as.vector(INC) @freakonometrics 13
  14. 14. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 7 > lig = as.factor(ligne); col = as.factor(colonne) 8 > 9 > CL <- glm(Y~lig+col , family= quasipoisson ) 10 > Y2=Y; Y2[is.na(Y)]=.001 11 > CL2 <- glm(Y2~lig+col , family= quasipoisson ) 12 > YP = predict(CL) 13 > p = 2*6-1; 14 > phi.P = sum(residuals(CL ,"pearson")^2)/(np -p) 15 > Sig = vcov(CL) 16 > X = model.matrix(CL2) 17 > Cov.eta = X%*%Sig%*%t(X) 18 > mu.hat = exp(predict(CL ,newdata=data.frame(lig ,col)))*futur 19 > pe2 = phi.P * sum(mu.hat) + t(mu.hat) %*% Cov.eta %*% mu.hat 20 > cat("Total reserve =", sum(mu.hat), "prediction error =", sqrt(pe2 ),"n") 21 Total reserve = 2426.985 prediction error = 131.7726 i.e. E [R − R]2 = 131.77. @freakonometrics 14
  15. 15. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Uncertainty and bootstrap simulations Based on that theoretical triangle, it is possible to generate residuals to obtain a simulated triangle. Since the size of the sample is small (here 21 observed values), assuming normality for Pearson’s residuals can be too restrictive. Resampling bootstrap procedure can then be more robust. In order to get the loss distribution, it is possible to use bootstrap techniques to generate a matrix of errors, see Renshaw & Verrall (1994). They suggest to boostrap Pearson’s residuals, and the simulation procedure is the following • estimate the model parameter (GLM), β, • calculate fitted values µi,j, and the residuals ri,j = Yi,j − µi,j V (µi,j) , • forecast with original data µi,j for i + j > n. Then can start the bootstrap loops, repeating B times @freakonometrics 15
  16. 16. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 • resample the residuals with resample, and get a new sample r (b) i,j , • create a pseudo sample solving Y ∗ i,j = µi,j + r (b) i,j × V (µi,j), • estimate the model using GLM procedure and derive boostrap forecast Let resid.sim be resampled residuals. Note that REG$fitted.values (called here payinc.pred) is the vector containing the µi,j’s. And further V (µi,j) is here simply REG$fitted.values since the variance function for the Poisson regression is the identity function. Hence, here Y ∗ i,j = µi,j + r (b) i,j × µi,j and thus, set 1 resid.sim = sample(resid ,Ntr*(Ntr +1)/2,replace=TRUE) 2 payinc.sim = resid.sim*sqrt(payinc.pred)+payinc.pred 3 4 [,1] [,2] [,3] [,4] [,5] [,6] @freakonometrics 16
  17. 17. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 5 [1,] 3155.699 1216.465 42.17691 18.22026 9.021844 22.89738 6 [2,] 3381.694 1245.399 84.02244 18.20322 11.122243 NA 7 [3,] 3726.151 1432.534 61.44170 23.43904 NA NA 8 [4,] 4337.279 1642.832 74.58658 NA NA NA 9 [5,] 4929.000 1879.777 NA NA NA NA 10 [6,] 5186.116 NA NA NA NA NA For this simulated triangle, we can use Chain-Ladder estimate to derive a simulated reserve amount (here 2448.175). Figure below shows the empirical distribution of those amounts based on 10, 000 random simulations. @freakonometrics 17
  18. 18. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 @freakonometrics 18
  19. 19. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Parametric or nonparametric Monte Carlo ? A natural idea would be to assume that Pearson residual have a Gaussian distribution, qqnorm(R); qqline(R) The graph on the right draw point with a size proportional to its Cook’s distance. @freakonometrics 19
  20. 20. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Instead of resampling in the sample obtained, we can also directly draw from a normal distribution, i.e. 1 > rnorm(length(R),mean=mean(R),sd=sd(R)) @freakonometrics 20
  21. 21. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 The second triangle is obtained using a Student t distribution (the blue line being the bootstrap estimate). 0.80 0.85 0.90 0.95 1.00 2400245025002550260026502700 VaR for total reserves probability level quantilelevel Student Normal bootstrap Note that the bootstrap technique is valid only in the case were the residuals are perfectly independent. @freakonometrics 21
  22. 22. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 In R, it is also possible to use the BootChainLadder(Triangle , R = 999, process.distr = "od.pois") function. @freakonometrics 22
  23. 23. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Going further So far, we have derived a ditrisbution for the best estimate of total reserves. Note tat it is possible to estimate a scale parameter φ. England & Verrall (1999) suggested φ = ε2 i,j n − p where the summation is over all past observations. @freakonometrics 23
  24. 24. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Bootstrap Chain-Ladder 1 > I=as.matrix(read.table("D: triangleC.csv",sep=";",header=FALSE)) 2 > BCL <- BootChainLadder (Triangle = I, R = 999, process.distr = "od. pois") 3 > BCL 4 BootChainLadder (Triangle = I, R = 999, process.distr = "od.pois") 5 6 Latest Mean Ultimate Mean IBNR SD IBNR IBNR 75% IBNR 95% 7 1 4 ,456 4 ,456 0.0 0.0 0 0 8 2 4 ,730 4 ,752 22.0 11.8 28 45 9 3 5 ,420 5 ,455 35.3 14.6 44 61 10 4 6 ,020 6 ,086 66.2 20.8 78 102 11 5 6 ,794 6 ,947 152.7 29.1 170 205 12 6 5 ,217 7 ,364 2 ,146.9 112.5 2 ,214 2 ,327 13 14 Totals 15 Latest: 32 ,637 16 Mean Ultimate: 35 ,060 @freakonometrics 24
  25. 25. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 17 Mean IBNR: 2 ,423 18 SD IBNR: 131 19 Total IBNR 75%: 2 ,501 20 Total IBNR 95%: 2 ,653 Histogram of Total.IBNR Total IBNR Frequency 2000 2200 2400 2600 2800 3000 0100200300 2000 2200 2400 2600 2800 3000 0.00.40.8 ecdf(Total.IBNR) Total IBNR Fn(x) q q qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qq qqqqqqq q q qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqq q q q q qq q qqq qq q qqqq q qq qqq q q 1 2 3 4 5 6 450060007500 Simulated ultimate claims cost origin period ultimateclaimscosts 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qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq q Mean ultimate claim qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq q q q qq qq q q q q qq qq q q q qq q q q q qqq qq qqq qq q q qq q qq q qq qqqq q qq qq q qqq qq qqqqq q qq qqqqqqqq q q qq q qq qq q qqqq q qq q qqq q qq qqqq q q qq qqq q q qq qq qq q qq qq qqq q q qq q q qq qqqq qqqqq q q qq q q q qqq q q qqqqqqqq q qq qqqq q qqqqq qqq qq qq qq q q qq qqqqqq q qqqq q q 1 2 3 4 5 6 020004000 Latest actual incremental claims against simulated values origin period latestincrementalclaims qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq Latest actual @freakonometrics 25
  26. 26. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 From Poisson to Over-Dispersed Poisson Classical, in GLMs we consider distributions with density f(z|θ, φ) = exp zθ − b(θ) φ + c(z, φ) , e.g. for the Poisson distribution P(λ) then f(z|λ) = exp(−λ) λz z! = exp z log λ − λ − log z! , z ∈ N, with θ = log λ, φ = 1, b(θ) = exp θ = λ and c(z, φ) = − log z!. Assume that φ = 1 becomes an additional parameter (that should be estimated). Note that in that case f(z|λ) is not any more a density, but it is a quasidensity. Further, note that V ar(Z) = φE(Z). Thus, if φ > 1 there is overdispersion. @freakonometrics 26
  27. 27. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 On quasiPoisson regression In order to understand the rule of the additional parameter, recall that for the Gaussien linear model, N(µ, σ2 ) it is an exponential distribution with θ = µ, b(θ) = θ2 /2, φ = σ2 and c(z, φ) = − 1 2 y2 σ2 + log(2πσ2 ) . Thus, φ is the variance parameter Y |X ∼ N(Xβ, σ2 ) In that linear model, estimation is done based on the following process, • estimate β as β = (X X)−1 X Y • derive the implied residuals, ε = Y − Xβ • estimate σ as the variance of the implied residuals @freakonometrics 27
  28. 28. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Thus, φ does not impact the estimation of the coefficient, but it will impact their significativity. 1 > an <- 10; ligne = rep (1:an , each=an); colonne = rep (1:an , an) 2 > passe = (ligne + colonne - 1) <=an; n = sum(passe) 3 > PAID=GenIns; INC=PAID 4 > INC[,2:an]= PAID [,2:an]-PAID [ ,1:(an -1)] 5 > Y = as.vector(INC) 6 > lig = as.factor(ligne) 7 > col = as.factor(colonne) 8 > base = data.frame(Y,col ,lig) 9 > reg1=glm(Y~col+lig ,data=base ,family="poisson") 10 > reg2=glm(Y~col+lig ,data=base ,family=" quasipoisson ") 11 > summary(reg1) 12 Call: 13 glm(formula = Y ~ col + lig , family = "poisson", data = base) 14 Coefficients : 15 Estimate Std. Error z value Pr(>|z|) 16 (Intercept) 12.5064047 0.0007540 16587.372 < 2e-16 *** @freakonometrics 28
  29. 29. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 17 col2 0.3312722 0.0006694 494.848 < 2e-16 *** 18 col3 0.3211186 0.0006877 466.963 < 2e -16 *** 19 col4 0.3059600 0.0007008 436.570 < 2e -16 *** 20 col5 0.2193163 0.0007324 299.461 < 2e-16 *** 21 col6 0.2700770 0.0007445 362.755 < 2e -16 *** 22 col7 0.3722084 0.0007606 489.344 < 2e -16 *** 23 col8 0.5533331 0.0008133 680.377 < 2e-16 *** 24 col9 0.3689342 0.0010429 353.772 < 2e-16 *** 25 col10 0.2420330 0.0018642 129.830 < 2e -16 *** 26 lig2 0.9125263 0.0006490 1406.042 < 2e-16 *** 27 lig3 0.9588306 0.0006652 1441.374 < 2e -16 *** 28 lig4 1.0259970 0.0006840 1499.927 < 2e-16 *** 29 lig5 0.4352762 0.0008019 542.814 < 2e -16 *** 30 lig6 0.0800565 0.0009364 85.492 < 2e -16 *** 31 lig7 -0.0063815 0.0010390 -6.142 8.14e-10 *** 32 lig8 -0.3944522 0.0013529 -291.560 < 2e -16 *** 33 lig9 0.0093782 0.0013963 6.716 1.86e -11 *** 34 lig10 -1.3799067 0.0039097 -352.946 < 2e -16 *** @freakonometrics 29
  30. 30. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 35 --- 36 ( Dispersion parameter for poisson family taken to be 1) 37 Null deviance: 10699464 on 54 degrees of freedom 38 Residual deviance: 1903014 on 36 degrees of freedom 39 (45 observations deleted due to missingness ) 40 AIC: 1903877 41 Number of Fisher Scoring iterations: 4 1 > summary(reg2) 2 Call: 3 glm(formula = Y ~ col + lig , family = " quasipoisson ", data = base) 4 Coefficients : 5 Estimate Std. Error t value Pr(>|t|) 6 (Intercept) 12.506405 0.172924 72.323 < 2e -16 *** 7 col2 0.331272 0.153537 2.158 0.03771 * 8 col3 0.321119 0.157719 2.036 0.04916 * 9 col4 0.305960 0.160736 1.903 0.06499 . 10 col5 0.219316 0.167970 1.306 0.19994 11 col6 0.270077 0.170756 1.582 0.12247 @freakonometrics 30
  31. 31. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 12 col7 0.372208 0.174451 2.134 0.03976 * 13 col8 0.553333 0.186525 2.967 0.00532 ** 14 col9 0.368934 0.239181 1.542 0.13170 15 col10 0.242033 0.427562 0.566 0.57485 16 lig2 0.912526 0.148850 6.131 4.65e-07 *** 17 lig3 0.958831 0.152569 6.285 2.90e-07 *** 18 lig4 1.025997 0.156883 6.540 1.33e-07 *** 19 lig5 0.435276 0.183914 2.367 0.02344 * 20 lig6 0.080057 0.214770 0.373 0.71152 21 lig7 -0.006381 0.238290 -0.027 0.97878 22 lig8 -0.394452 0.310289 -1.271 0.21180 23 lig9 0.009378 0.320249 0.029 0.97680 24 lig10 -1.379907 0.896690 -1.539 0.13258 25 --- 26 ( Dispersion parameter for quasipoisson family taken to be 52601.93) 27 Null deviance: 10699464 on 54 degrees of freedom 28 Residual deviance: 1903014 on 36 degrees of freedom 29 (45 observations deleted due to missingness ) @freakonometrics 31
  32. 32. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 30 AIC: NA 31 Number of Fisher Scoring iterations: 4 Thus, coefficients are identical so it not affect the best estimate of claims reserves.... unless we take into account the fact that some variates are no longer significant..... 1 > base = data.frame(Y,col ,lig) 2 > base$lig[base$lig=="7"]="1" 3 > base$lig[base$lig=="9"]="1" 4 > base$lig[base$lig=="6"]="1" 5 > base$col[base$col=="5"]="1" 6 > base$col[base$col=="10"]="1" 7 > base$col[base$col=="9"]="1" 8 > base$col[base$col=="6"]="1" 9 > base$col[base$col=="4"]="1" 10 > base$col[base$col=="3"]="1" 11 > base$col[base$col=="7"]="1" 12 > base$col[base$col=="2"]="1" @freakonometrics 32
  33. 33. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 13 > base$lig[base$lig=="8"]="1" 14 > summary(glm(Y~col+lig ,data=base ,family=" quasipoisson ")) 15 Call: 16 glm(formula = Y ~ col + lig , family = " quasipoisson ", data = base) 17 Coefficients : 18 Estimate Std. Error t value Pr(>|t|) 19 (Intercept) 12.73401 0.07764 164.022 < 2e-16 *** 20 col8 0.28877 0.14109 2.047 0.04618 * 21 lig2 0.96246 0.10984 8.763 1.59e -11 *** 22 lig3 0.99721 0.11232 8.878 1.07e -11 *** 23 lig4 1.06465 0.11481 9.273 2.82e -12 *** 24 lig5 0.45513 0.14622 3.113 0.00312 ** 25 lig10 -1.60752 0.85482 -1.881 0.06611 . 26 --- 27 ( Dispersion parameter for quasipoisson family taken to be 49241.53) 28 Null deviance: 10699464 on 54 degrees of freedom 29 Residual deviance: 2442092 on 48 degrees of freedom 30 (45 observations deleted due to missingness ) @freakonometrics 33
  34. 34. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 31 AIC: NA 32 Number of Fisher Scoring iterations: 4 Thus, 1 > M= cbind(Y,predict(reg1 ,newdata=base0 ,type="response"), 2 + predict(reg2 ,newdata=base0 ,type="response"), 3 + predict(reg3 ,newdata=base , type="response")) 4 > sum(M[is.na(Y)==TRUE ,2]) 5 [1] 18680856 6 > sum(M[is.na(Y)==TRUE ,3]) 7 [1] 18680856 8 > sum(M[is.na(Y)==TRUE ,4]) 9 [1] 18226919 Including an overdispersion parameter φ might impact the estimation of the overall reserves. @freakonometrics 34
  35. 35. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Testing for overdispersion In order to test for overdispersion in an econometric model, we need to specify how overdispersion appears. A classical test is to assume that V ar(Y |X) = E(Y |X) + τE(Y |X)2 which is a standard econometric model with random effect. We want to test H0 : τ = 0 against H1 : τ > 0 A standard test statistics is T = n i=1[Yi − λi]2 − Yi 2 n i=1 λ2 i which has a N(0, 1) distribution under H0. An alternative is to consider T = n i=1[Yi − λi]2 − Yi n i=1[[Yi − λi]2 − Yi]2 @freakonometrics 35
  36. 36. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Those test can be found in R, respectively 1 > library(AER) 2 > dispersiontest (reglmp) 3 > dispersiontest (reglmp ,trafo = 2) An alternative is simply the following 1 > library(ChainLadder) 2 > an <- 10; ligne = rep (1:an , each=an); colonne = rep (1:an , an) 3 > passe = (ligne + colonne - 1) <=an; n = sum(passe) 4 > PAID=GenIns; INC=PAID 5 > INC [,2:an]= PAID [,2:an]-PAID [ ,1:(an -1)] 6 > Y = as.vector(INC) 7 > lig = as.factor(ligne) 8 > col = as.factor(colonne) 9 > base = data.frame(Y,col ,lig) 10 > reg1=glm(Y~col+lig ,data=base ,family="poisson") 11 > reg2=glm(Y~col+lig ,data=base ,family=" quasipoisson ") 12 > dispersiontest (reg1) @freakonometrics 36
  37. 37. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 13 Overdispersion test 14 data: reg1 15 z = 4.3942 , p-value = 5.558e -06 16 alternative hypothesis : true dispersion is greater than 1 @freakonometrics 37
  38. 38. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Alternative models for overdispersion There is overdispersion if Var(Y ) > E(Y ), which can be obtained with a negative binomial distribution (with belongs to the exponential family) 1 > library(MASS) 2 > reg3=glm.nb(Y~col+lig ,data=base) 3 > summary(reg3) 4 ( Dispersion parameter for Negative Binomial (13.8349) family taken to be 1) 5 Theta: 13.83 6 Std. Err.: 2.61 7 2 x log -likelihood: -1460.766 8 > sum(exp(predict(reg3 ,newdata=base))[passe!=TRUE ]) 9 [1] 18085795 @freakonometrics 38
  39. 39. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Uncertainty and overdispersion Based on the explicit expression for the prediction error, it is possible to obtain prediction error for those three models, 1 > predCL=function(reg=reg1 ,regb=reg1b){ 2 + p = 2*6-1; 3 + phi.P = sum(residuals(reg ,"pearson")^2)/(np -p) 4 + Sig = vcov(reg) 5 + X = model.matrix(regb) 6 + Cov.eta = X%*%Sig%*%t(X) 7 + mu.hat = exp(predict(reg ,newdata=data.frame(lig ,col)))*futur 8 + pe2 = phi.P * sum(mu.hat) + t(mu.hat) %*% Cov.eta %*% mu.hat 9 + cat("Total reserve =", sum(mu.hat), " prediction error =", sqrt(pe2 ),sqrt(pe2)/sum(mu.hat),"n") 10 + } Avec nos trois modèles, Poisson, ODP et binomiale négative, on obtient, 1 > predCL(reg1 ,reg1b) @freakonometrics 39
  40. 40. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 2 Total reserve = 18680856 prediction error = 896876.9 0.04801048 3 > predCL(reg2 ,reg2b) 4 Total reserve = 18680856 prediction error = 4736425 0.2535443 5 > predCL(reg3 ,reg3b) 6 Total reserve = 18085795 prediction error = 2058134 0.1137984 @freakonometrics 40
  41. 41. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 On the prediction error In order to derive an estimation of the prediction error using bootstrap techniques, we have not only to generate randomly possible triangles, but also to add uncertainty in the developpement, using e.g. the fact that Ci,j+1 = λjCi,j + σj Ci,j + εi,j where the noise can be assume to be Gaussian, N(0, 1). The statistical interpretation is that Ci,j+1|Ci,j ∼ N(λjCi,j + σ2 j Ci,j) Classically we use 1 > CL=function(triangle){ 2 + n=nrow(triangle) 3 + LAMBDA=rep(NA ,n -1) 4 + for(i in 1:(n-1)){ @freakonometrics 41
  42. 42. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 5 + LAMBDA[i]= sum(triangle [1:(n-i),i+1])/ 6 + sum(triangle [1:(n-i),i]) } 7 + DIAG=diag(triangle[,n:1]) 8 + TOTO=c(1,rev(LAMBDA)) 9 + return(sum(cumprod(TOTO)*DIAG -DIAG)) } a natural idea is to consider 1 > CLboot=function(triangle ,l,s){ 2 + m=nrow(triangle) 3 + for(i in 2:m){ 4 + triangle [(m-i+2):m,i]= rnorm(i-1, 5 + mean=triangle [(m-i+2):m,i-1]*l[i-1], 6 + sd=sqrt(triangle [(m-i+2):m,i -1])*s[i -1]) 7 + } 8 + ULT=triangle[,m] 9 + DIAG=diag(triangle[,m:1]) 10 + return(sum(ULT -DIAG)) } Then, we can run boostrap simulations, @freakonometrics 42
  43. 43. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 1 > base=data.frame(Y,lig ,col) 2 > REG=glm(Y~lig+col ,family=poisson) 3 > YP=predict(REG ,newdata=base) 4 > E=residuals(REG ,"pearson") 5 > PROV.BE=rep(NA ,5000) 6 > PROVISION=rep(NA ,5000) 7 > for(k in 1:50000){ 8 + simE=sample(E,size =36, replace=TRUE) 9 + bruit=simE*sqrt(exp(YP)) 10 + INCsim=exp(YP)+bruit 11 + INCM=matrix(INCsim ,6 ,6) 12 + CUMM=INCM 13 + for(j in 2:6){CUMM[,j]= CUMM[,j -1]+ INCM[,j]} 14 + PROV.BE[k]=CL(CUMM) 15 + PROVISION[k]= CLboot(CUMM ,lambda ,sigma)} @freakonometrics 43
  44. 44. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Random Generation of a Quasi-Distribution It is also possible to generate Poisson, or quasi-Poisson random variables. Recall that the negative binomial distribution has probability function P[N = k] = Γ(k + r) k!Γ(r) · [1 − p]r pk where the expected value and the variance are µ = r · p 1 − p and σ2 = µ = r · p (1 − p)2 Assume that σ2 = ϕ · µ, then r = µ ϕ − 1 and p = 1 ϕ 1 > rqpois = function(n, lambda , phi) { 2 + return( rnbinom(n, size = lambda/(1-phi), prob = 1/phi) } @freakonometrics 44
  45. 45. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Using GAM for claims reserving In the case of GAM’s, assume that Yi,j ∼ L(θçi, j), where θi,j = ϕ(u(i) + v(j)), where here u and v are two unknown functions. We still have an additive form, but on unknown transformations of explanatory variates. Spline functions are considered to estimation functions u and v. 1 > library(gam) 2 > GAM=gam(payinc~s(year ,5)+s(development ,3),data=D,familly="Poisson") 3 > plot.gam(GAM ,se=T,col="red",ask=TRUE ,main="GAM model , df=5, df=3") @freakonometrics 45
  46. 46. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 @freakonometrics 46
  47. 47. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Dealing with negative increments Negative incremental values can arise due to timing of reinsurance, recoveries, cancellation of outstanding claims. One might argue that the problem is more with the data than with the methods. England & Verall (2002) mention that the Gaussian model is less affected by the presence of negative incremental values. Unfortunately, one can hardly assume that data are Gaussian because of the skewness. Renshaw & Verral (1994) suggested to add a “small constant” to the past data and to substract this constant from forecasts at the end. @freakonometrics 47
  48. 48. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Dealing with negative increments A classical technique to avoid negative payments is to consider a translation of the incremental triangle, i.e. Y + i,j = Yi,j + κ such that Y + i,j > 0 for all i, j. @freakonometrics 48
  49. 49. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Dealing with negative increments A classical technique to avoid negative payments is to consider a translation of the incremental triangle, i.e. Y + i,j = Yi,j + κ such that Y + i,j > 0 for all i, j. q q q 0 1 2 3 4 012345 q q q @freakonometrics 49
  50. 50. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Why a Poisson regression model ? The is no reason to assume that incremental payments are Poisson distribution. The only motivation here is that the expected value is the same as the Chain Ladder estimate. Distribution of the reserves, GAM model with Chain Ladder Total amount of reserves Density 2000 4000 6000 8000 0e+001e−042e−043e−044e−045e−04 @freakonometrics 50
  51. 51. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Tweedie ? The density of a tweedie model with power function p would be 1 > ftweedie = function(y,p,mu ,psi){ 2 + if(p==2){f = dgamma(y, 1/psi , 1/(psi*mu))} else 3 + if(p==1){f = dpois(y/psi , mu/psi)} else 4 + {lambda = mu^(2-p)/psi /(2-p) 5 + if(y==0){ f = exp(-lambda)} else 6 + { alpha = (2-p)/(p -1) 7 + beta = 1 / (psi * (p -1) * mu^(p -1)) 8 + k = max (10, ceiling(lambda + 7*sqrt(lambda))) 9 + f = sum(dpois (1:k,lambda) * dgamma(y,alpha*(1:k),beta)) 10 + }} 11 + return(f) 12 + } A numerical problem is that we should have no missing values in the regression, so artificially, consider 1 > source("http:// freakonometrics .free.fr/bases.R") @freakonometrics 51
  52. 52. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 2 > library(statmod) 3 > an <- 6; ligne = rep (1:an , each=an); colonne = rep (1:an , an) 4 > passe = (ligne + colonne - 1) <=an; n = sum(passe) 5 > INC=PAID 6 > INC [ ,2:6]= PAID [,2:6]- PAID [ ,1:5] 7 > Y = as.vector(INC) 8 > lig = as.factor(ligne) 9 > col = as.factor(colonne) 10 > y = Y[passe] 11 > Y[is.na(Y)]=.01 Then, we can run an econometric regression 1 > pltweedie <- function(pow){ 2 + regt = glm(Y~lig+col , tweedie(pow ,0)) 3 + reserve = sum(fitted.values(regt)[!passe ]) 4 + dev = deviance(regt) 5 + phi.hat = dev/n 6 + mu = fitted.values(regt)[passe] @freakonometrics 52
  53. 53. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 7 + hat.logL = 0 8 + for (k in 1: length(y)){ 9 + hat.logL <- hat.logL + log(ftweedie(y[k], pow , mu[k], phi.hat)) } 10 + cat("Puissance =", round(pow ,3) , "phi =", round(phi.hat ,2), 11 + "Reserve (tot) =", round(reserve), "logL =", round(hat.logL ,3)) 12 + hat.logL} 13 > for(pow in c(1 ,1.25 ,1.5 ,1.75 ,2)){pltweedie(pow)} 14 Puissance = 1 phi = 166.95 Reserve (tot) = 1345 logL = -Inf 15 Puissance = 1.25 phi = 42.92 Reserve (tot) = 1216 logL = -151.72 16 Puissance = 1.5 phi = 15.8 Reserve (tot) = 996 logL = -145.232 17 Puissance = 1.75 phi = 9.02 Reserve (tot) = 609 logL = -153.997 18 Puissance = 2 phi = 6.78 Reserve (tot) = 125 logL = -170.614 It is also possible to run a optimization routine, 1 > optimize(pltweedie , c(1.01 ,1.99) , tol =1e-4, maximum = TRUE) 2 -144.624 3 $maximum @freakonometrics 53
  54. 54. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 4 [1] 1.427873 5 6 $objective 7 [1] -144.6237 Thus, here the Poisson model might not be the appropriate one, 1.0 1.2 1.4 1.6 1.8 2.0 −2000−1500−1000−500 power Loglikelihood @freakonometrics 54
  55. 55. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Bayesian Models in Claims Reserving The first idea is to consider some credibility based model, with Ci,n = Z · CMack i,n + [1 − Z] · µi given some a priori µi. For instance Benkhtander (1976) and Hovinen (1981) suggested Z = 1 − [1 − βi]2 where βi = n−1 k=n−i 1 λk Note that Ci,n = Ci,n−i + [1 − βi] βi · ·CMack i,n + [1 − βi] · µi More generally, consider the Cape-Code technique, Ci,n = Ci,n−i + 1 − Ci,n−i Ci,n Ci,n @freakonometrics 55
  56. 56. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 sous la forme Ci,n = Ci,n−i + 1 − Ci,n−i Ci,n LRi · Pi, où LRi correspond au loss ratio pour l’année i, i.e. LRi = Ci,n/Pi. L’idée de la méthode dite Cape-Code est d’écrire une forme plus générale, Ci,n = Ci,n−i + (1 − πn−i) LRiPi où πn−i correspond à une cadence de paiement, et peut être estimé par la méthode Chain Ladder. Quant aux LRi il s’agit des loss ratio cibles, correspondant à un avis d’expert. On peut aussi proposer un même ratio cible pour plusieurs années de survenance. On posera alors Ri = Ci,n − Ci,n−i = (1 − πn−i)LRAPi. pour i ∈ A, où LRA = k∈A Cn,n−k k∈A πn−kPk . @freakonometrics 56
  57. 57. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Dans un premier temps, on peut calculer les πi à partir de la méthode Chain Ladder, i.e. πn−i = Ci,n−i Ci,n où la charge ultime est celle prédite pas la méthode Chain-Ladder. 1 > Cultime = MackChainLadder (PAID)$ FullTriangle [,6] 2 > (PI <- (1- Cdiag/Cultime)) 3 1 2 3 4 5 6 4 0.00000 0.00471 0.00656 0.01086 0.02204 0.29181 5 > LR <- TRIANGLE [,6]/PREMIUM 6 > Cdiag <- diag(PAID [ ,6:1]) 7 > (Cultime -Cdiag)/(LR*PREMIUM) 8 1 2 3 4 5 6 9 0.00000 0.00471 0.00656 0.01086 0.02204 0.29181 Si on suppose ensuite que A = {1, 2, · · · , n}, alors 1 > LR=sum(TRIANGLE [ ,6])/sum(PREMIUM) @freakonometrics 57
  58. 58. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 2 > PI*LR*PREMIUM 3 1 2 3 4 5 6 4 0.0 24.6 35.6 62.7 139.6 2095.3 5 > sum(PI*LR*PREMIUM) 6 [1] 2358 On obtient ici un montant de provision total inférieur à celui obtenu par la méthode Chain Ladder puisque le montant de provisions vaut ici 2357.756. @freakonometrics 58
  59. 59. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Modèles bayésiens et Chain Ladder De manière générale, un méthode bayésienne repose sur deux hypothèses • une loi a priori pour les paramètres du modèle (Xi,j, Ci,j, λi,j, LRi,j = Ci,j/Pj, etc) • une technique pour calculer les lois a posteriori, qui sont en général assez complexes. @freakonometrics 59
  60. 60. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Modèles bayésiens pour les nombres de sinistres Soit Ni,j l’incrément du nombre de sinistres, i.e. le nombre de sinistres survenus l’année i, déclarés l’année i + j. On note Mi le nombre total de sinistres par année de survenance, i.e. Mi = Ni,0 + Ni,1 + · · · . Supposons que Mi ∼ P(λi), et que p = (p0, p1, · · · , pn) désigne les proprotions des paiments par année de déroulé. Conditionnellement à Mi = mi, les années de survenance sont indépenantes, et le vecteur du nombre de sinistres survenus année l’année i suit une loi multinomiale M(mi, p). @freakonometrics 60
  61. 61. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Modèles bayésiens pour les nombres de sinistres La vraisemblance L(M0, M1, · · · , Mn, p|Ni,j) est alors n i=0 Mi! (Mi − Nn−i)!Ni,0!Ni,1! · · · Ni,n−i! [1 − pn−i]Mi−Nn−i p Ni,0 0 p Ni,1 1 · · · p Ni,n−i n−i où Nn−i = N0 + N1 + · · · + Nn−i et pn−i = p0 + p1 + · · · + pn−i. Il faut ensuite de donner une loi a priori pour les paramètres. La loi a posteriori sera alors proportionnelle produit entre la vraisemblance et cette loi a priori. @freakonometrics 61
  62. 62. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Modèles bayésiens pour les montants agrégés On pose Yi,j = log(Ci,j), et on suppose que Yi,j = µ + αi + βj + εi,j, où εi,j ∼ N(0, σ2 ). Aussi, Yi,j suit une loi normale, f(yi,j|µ, α, β, σ2 ) ∝ 1 σ exp − 1 2σ2 [yi,j − µ − αi − βj] 2 , et la vraisemblance est alors L(θ, σ|Y ) ∝ σ−m exp   i,j [yi,j − µ − αi − βj] 2   où m = (n(n + 1)/2 désigne le nombre d’observations passées. La difficulté est alors de spécifier une loi a priori pour (θ, σ2 ), i.e. (µ, α, β, σ2 ). @freakonometrics 62
  63. 63. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 MCMC and Bayesian Models We have a sample x = {x1, · · · , xd) i.i.d. from distribution fθ(·). In predictive modeling, we need E(g(X)|x) = g(x)fθ|x(x)dx where fθ|x(x) = f(x|x) = f(x|θ) · π(θ|x)dθ How can we derive π(θ|x) ? Can we sample from π(θ|x) (and use monte carlo technique to approximate the integral) ? @freakonometrics 63
  64. 64. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Hastings-Metropolis Back to our problem, we want to sample from π(θ|x) i.e. generate θ1, · · · , θn, · · · from π(θ|x). Hastings-Metropolis sampler will generate a Markov Chain (θt) as follows, • generate θ1 • generate θ and U ∼ U([0, 1]), compute R = π(θ |x) π(θt|x) P(θt|θ ) P(θ |θt−1) if U < R set θt+1 = θ if U ≥ R set θt+1 = θt R is the acceptance ratio, we accept the new state θ with probability min{1, R}. @freakonometrics 64
  65. 65. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Hastings-Metropolis Observe that R = π(θ ) · f(x|θ ) π(θt) · f(x|θt) P(θt|θ ) P(θ |θt−1) In a more general case, we can have a Markov process, not a Markov chain. E.g. P(θ |θt) ∼ N(θt, 1) @freakonometrics 65
  66. 66. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Using MCMC to generate Gaussian values 1 > metrop1 <- function(n=1000 , eps =0.5){ 2 + vec <- vector("numeric", n) 3 + x=0 4 + vec [1] <- x 5 + for (i in 2:n) { 6 + innov <- runif(1,-eps ,eps) 7 + mov <- x+innov 8 + aprob <- min(1, dnorm(mov)/dnorm(x)) 9 + u <- runif (1) 10 + if (u < aprob) 11 + x <- mov 12 + vec[i] <- x 13 + } 14 + return(vec)} @freakonometrics 66 q q −2 −1 0 1 2 q
  67. 67. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Using MCMC to generate Gaussian values 1 > plot.mcmc <- function(mcmc.out){ 2 + op <- par(mfrow=c(2 ,2)) 3 + plot(ts(mcmc.out),col="red") 4 + hist(mcmc.out ,30, probability =TRUE , 5 + col="light blue") 6 + lines(seq(-4,4,by =.01) ,dnorm(seq (-4,4, 7 + by =.01)),col="red") 8 + qqnorm(mcmc.out) 9 + abline(a=mean(mcmc.out),b=sd(mcmc. out)) 10 + acf(mcmc.out ,col="blue",lag.max =100) 11 + par(op)} 12 > metrop.out <-metrop1 (10000 ,1) 13 > plot.mcmc(metrop.out) @freakonometrics 67 Time ts(mcmc.out) 0 2000 4000 6000 8000 10000 −3−2−10123 Histogram of mcmc.out mcmc.out Density −4 −3 −2 −1 0 1 2 3 0.00.10.20.30.4 qq q q q qq qq qq q q q qq q qq q q q q q q q q qqqq qq q q q q q qq qq q q q q q q q q q qq q q q q q q q q q q q q qqq q qq q q q q q q q q q q q qq q q qqqqq q qqqqq q q q q qqqqq qq q qq q q q q qq q qq qq q q q q q q q q qqq q q q qq q qq q q q q q q q qq qq q q qqq q q q q qq qqqqq q q qq qq q q q q q q q q q qq q q q q q qq q q q q q qq qq q q q q q qqq q q qq q q qq qq q q qq q q qq q q qq q qq q q qqq q q q qq qq q q q q q q qqqqq qq q q q q q q q q q q q q q q q q q q q q q q q q q qq q q qq q q q q q q q qq q q qqq q q qq q q qq q q q qq q q q q q q qq qq qqqqqq q qq q q q qq q q q q qq q q q q q q q q q q q q qqq q q q qq q q qq qq q qq q q q q qq q qqq q q q qq q q qqqq q q q q q q qqq q q qq q qqq q q qqq q q qqq q q q q qq qq qq qq qqq q qq q qq qq qq q q q q q q qqq q q q q q q q q q q q q q q q q qq q qq q q q q q q qq q q q q q q q q qq q q q q q q q q q q qqq qqq q q q qqq q qq q q q q qq q q qq q q q qqqq q qqqq q q q q q q q qq q q q qqq q q q q q q qq qqq q q qq q qq qq qq q q q q qqqq q q q q q q qq qq q q q q q qq qq qq q q qqq q q q q q q q q q q q qqq q q q q q q q q q q qqq q q q q q qq q q q q q q q q q q q q q q q q q q qq q q q q q q q q qqq q q qqq q q q q q q qq q qq q qq q q q q q q q q q q q q q qqq q q qq q qq q q q q qq q qq qq qq q q qq q q q q qqq q q q q qq qqqq q q 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qq qq q q q q q qq qq q q qqq q qqq q q qq q q qq q q q qq q q q qqq q q qq q q q qq q qq q q q q q q q q q q q qq q qq q q q q q qq q qq q q q q q q q qq qq qq q q q q q q qqq qq qq q qq q q q q qq q q q qq qq q q q q q q q qq q q q q q q q q qq qq q qq q q q q q q q q qq qq q qq q q q qq q q q qqqq q qq qq q qq q qqq qq q q q q q qq q qqq q qq qq q q q q q q q q q q q q q q q q q qqqq qqqqq qqq q qq qq q q q q qq q qq q qq q q q q q q q q q q q q qq qq q qq q q qq q qq q q q q q qq q q q q q q q q qqqq q qqq q q q q q q qq q qq q qq q qq q q qqqq q q q q q qq qq q q qq q q q q q qqq qq q q q q q qq q q q qq qq q q q qq qqq q qqq qq q q q qqq q q q qq q q qqq q qq q q qq q q qq qq qqq qq q qqq q q q q q qq q q qq q q qqq q q q q q q q q q qq qq qq q q q q q q q q qq q q q qq q q q q q q q q qq q q q q qqq q q q q qq q q q q q q q qq qq q q q q q q q q q q q q q qq q q q q q q q q q q qq q qqqq q q qqq qq q q q q q qq q q qqqq q q q q q qq qq q q q qq q q q q q q q q q qq q q q q q q q q q qqq q q qq q q q q qq q qq qq q q q qq q q qq q q q q q q qqqq q q qq q q q q q q q qq q q q q q q q q q −4 −2 0 2 4 −3−2−10123 Normal Q−Q Plot Theoretical Quantiles SampleQuantiles 0 20 40 60 80 100 0.00.20.40.60.81.0 Lag ACF Series mcmc.out
  68. 68. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Heuristics on Hastings-Metropolis In standard Monte Carlo, generate θi’s i.i.d., then 1 n n i=1 g(θi) → E[g(θ)] = g(θ)π(θ)dθ (strong law of large numbers). Well-behaved Markov Chains (P aperiodic, irreducible, positive recurrent) can satisfy some ergodic property, similar to that LLN. More precisely, • P has a unique stationary distribution λ, i.e. λ = λ × P • ergodic theorem 1 n n i=1 g(θi) → g(θ)λ(θ)dθ even if θi’s are not independent. @freakonometrics 68
  69. 69. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Heuristics on Hastings-Metropolis Remark The conditions mentioned above are • aperiodic, the chain does not regularly return to any state in multiples of some k. • irreducible, the state can go from any state to any other state in some finite number of steps • positively recurrent, the chain will return to any particular state with probability 1, and finite expected return time @freakonometrics 69
  70. 70. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 MCMC and Loss Models Example A Tweedie model, E(X) = µ and Var(X) = ϕ · µp . Here assume that ϕ and p are given, and µ is the unknown parameter. → need a predictive distribution for µ given x. Consider the following transition kernel (a Gamma distribution) µ|µt ∼ G µt α , α with E(µ|µt) = µt and CV(µ) = 1 √ α . Use some a priori distribution, e.g. G (α0, β0). @freakonometrics 70
  71. 71. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 MCMC and Loss Models • generate µ1 • at step t : generate µ ∼ G α−1 µt, α and U ∼ U([0, 1]), compute R = π(µ ) · f(x|µ ) π(µt) · f(x|θt) Pα(µt|θ ) Pα(θ |θt−1) if U < R set θt+1 = θ if U ≥ R set θt+1 = θt where f(x|µ) = L(µ) = n i=1 f(xi|µ, p, ϕ), f(x · |µ, p, ϕ) being the density of the Tweedie distribution, dtweedie function (x, p, mu, phi) from library(tweedie). @freakonometrics 71
  72. 72. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 1 > p=2 ; phi=2/5 2 > set.seed (1) ; X <- rtweedie (50,p,10,phi) 3 > metrop2 <- function(n=10000 , a0=10, 4 + b0=1, alpha =1){ 5 + vec <- vector("numeric", n) 6 + vec [1] <- rgamma (1,a0 ,b0) 7 + for (i in 2:n){ 8 + mustar <- rgamma (1,vec[i -1]/alpha ,alpha) 9 + R=prod(dtweedie(X,p,mustar ,phi)/dtweedie 10 + (X,p,vec[i-1],phi))*dgamma(mustar ,a0 ,b0) 11 + dgamma(vec[i-1],a0 ,b0)* dgamma(vec[i-1], 12 + mustar/alpha ,alpha)/dgamma(mustar , 13 + vec[i-1]/alpha ,alpha) 14 + aprob <- min(1,R) 15 + ifelse(runif (1) < aprob ,vec[i]<-mustar , 16 + vec[i]<-vec[i -1])} 17 + return(vec)} 18 > metrop.output <-metrop2 (10000 , alpha =1) @freakonometrics 72 Time ts(mcmc.out) 0 2000 4000 6000 8000 10000 789101112 Histogram of mcmc.out mcmc.out Density 7 8 9 10 11 12 0.00.10.20.30.40.5 qq qq q qq q qqqqqqq qqq qqqqqqqqq qqqqqqqq qqq qqqq qqqqqqqqqqq qqqqqqqqqqqqq q qqq qqqq qqqqq q q qqqq q q q qqqqqq qq qq qqq qqqqqqq qqqq qqqqqq 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  73. 73. Arthur CHARPENTIER - Actuariat de l’Assurance Non-Vie, # 11 Gibbs Sampler For a multivariate problem, it is possible to use Gibbs sampler. Example Assume that the loss ratio of a company has a lognormal distribution, LN(µ, σ2 ), .e.g 1 > LR <- c(0.958 , 0.614 , 0.977 , 0.921 , 0.756) Example Assume that we have a sample x from a N(µ, σ2 ). We want the posterior distribution of θ = (µ, σ2 ) given x . Observe here that if priors are Gaussian N µ0, τ2 and the inverse Gamma distribution IG(a, b), then    µ|σ2 , x ∼ N σ2 σ2 + nτ2 µ0 + nτ2 σ2 + nτ2 x, σ2 τ2 σ2 + nτ2 2 i=1 σ2 |µ, x ∼ IG n 2 + a, 1 2 n i=1 [xi − µ]2 + b More generally, we need the conditional distribution of θk|θ−k, x, for all k. 1 > x <- log(LR) @freakonometrics 73

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